A $μ -$ -meson particle of charge equal to that of an electron, $- e$ and mass 208 times the mass of the electron moves in a circular orbit around the nucleus of charge $+ 4 e$ . Assuming the Bohrs model of the atom to be applicable to this system, the inverse of wavelength of photons emitted when an electron jumps in the 4th orbit of the atom from infinity in terms of Rydberg constant is given by $α R_{H}$ . Find the value of $α / 26$ .

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Solution

For electron , $\frac{1}{λ} = R_{H} Z^{2} [1 / n_{1}^{2} - 1 / n_{2}^{2}]$
As $E \propto m$ so for meson Rydberg constant will be $208 R_{H}$
and given Z=4$
For meson, $\frac{1}{λ} = 208 R_{H} (4)^{2} [1 / 4^{2} - 1 / \infty^{2}] = 208 R_{H}$
Now, $α R_{H} = 208 R_{H}$ or $α = 208$
Thus, $α / 26 = 208 / 26 = 8$

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